Additional thought to otherwise appropriate answer: if you don't have a calculator and want a numerical estimate, it's probably better to look at the base-10 logarithm of this.
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orionMay 7 '14 at 13:27

11

Any equation that includes both e and pi always looks like witchcraft to me.
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mikeTheLiarMay 7 '14 at 17:36

2

@mikeTheLiar Study complex analysis. There is no end to the witchcraft in that field.
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6005May 7 '14 at 19:25

3

@Goos Alternatively, there is an immediate end to the witchcraft once you understand the relationship between exponents and angles.
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Kyle StrandMay 7 '14 at 21:59

which I think is the right place to start for any answer to this question. However, it may be useful to add that if you take the logarithm of this amount, you get the (approximate) number of digits you need to write down $n!$, which may be of more use:

You could of course use logs to do multiplication </oldfogey> In any case it becomes $\log_{10}(n!) \sim n\log_{10}(n) - n \log_{10}(e) + \frac{1}{6}\log_{10}(n(1+4n(1+2n))) + \frac{1}{2}\log_{10}(\pi).$
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HenryMay 8 '14 at 9:45

A different tack here, as you I suspect you are asking for a method that doesn't require any 'advanced' functions.

How large a positive integer number is, can be expressed well in terms of how many digits it takes to write the number. This is best represented as the integer part of the logarithm of the number, plus 1 (logarithm base 10, i.e. Log not Ln). For example LOG[55] = 1.74 (2 d.p.), it requires INT[1.74] + 1 = 2 digits to represent it. You can get a better feel for the size of a number by looking at the digits after the decimal point in the logarithm too.

Yes, i Was looking for something like this. I Was surprised though, by the variety of answers... So i'm checking them aswell :)
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TruxMay 8 '14 at 4:28

An addition, I perhaps wasn't clear what the error factor was. This is the error in n!. So n! is within a factor of 2 across the full 120 to 1000 range, LOG[n!] is within +/-0.29.
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user1228123May 8 '14 at 9:37